Solved: A botanist conducted a nine-week experiment on two plants, $X$ and $Y$, of the... [IB Mathematics Analysis and Approaches (AA)]

Question

SLPaper 2

A botanist conducted a nine-week experiment on two plants, $X$ and $Y$ , of the same species. She wanted to determine the effect of using a new plant fertilizer. Plant $X$ was given fertilizer regularly, while Plant $Y$ was not.

The botanist found that the height of Plant $X$ , at time $t$ weeks can be modelled by the function $h_X(t) = \sin(2t+6)+9t+27$ , where $0 \leq t \leq 9$ .

The botanist found that the height of Plant $Y$ , at time $t$ weeks can be modelled by the function $h_Y(t) = 8t+32$ , where $0 \leq t \leq 9$ .

Use the botanist's models to find the initial height of

Plant $X$ correct to three significant figures.

[2]

Verified

Solution

$h_X(0) = \sin(6) + 27$ M1
$= 26.7205$
$= 26.7$ (cm) A1

2 marks total

Remember to include units (cm) in your final answer

Plant $Y$ .

[1]

Verified

Solution

Initial height of Plant Y is when $t = 0$ :

$h_Y(0) = 8(0) + 32 = 32$ A1

1 mark total

Remember that the initial height corresponds to t = 0 in the given function

Find the values of $t$ when $h_X(t) = h_Y(t)$ .

[3]

Verified

Solution

Attempts to solve $h_X(t) = h_Y(t)$ for $t$ M1
Equation to solve: $\sin(2t+6)+9t+27 = 8t+32$
Solution: $t = 4.00746..., 4.70343..., 5.88332...$
Final answer: $t = 4.01, 4.70, 5.88$ (weeks) A2

3 marks total

Award A1 for two correct solutions and A2 for all three correct solutions

For $0 \leq t \leq 9$ , find the total amount of time when the rate of growth of Plant $Y$ was greater than the rate of growth of Plant $X$ .

[6]

Verified

Solution

Recognizes that $h_X'(t)$ and $h_Y'(t)$ are required M1
Attempts to solve $h_X'(t) = h_Y'(t)$ for $t$ M1
$t = 1.18879$ and $2.23598$ OR $4.33038$ and $5.37758$ OR $7.47197$ and $8.51917$ A1

Award full marks for $t=\frac{4\pi}{3}-3,\ \frac{5\pi}{3}-3,\ \frac{7\pi}{3}-3,\ \frac{8\pi}{3}-3,\ \frac{10\pi}{3}-3,\ \frac{11\pi}{3}-3$ .

Award subsequent marks for correct use of these exact values.

$1.18879 < t < 2.23598$ OR $4.33038 < t < 5.37758$ OR $7.47197 < t < 8.51917$ A1
Attempts to calculate the total amount of time M1
$3(2.2359...-1.1887...)$

$=3\left(\frac{5\pi}{3}-3-\left(\frac{4\pi}{3}-3\right)\right)$

$=3.14\ (\text{weeks})$ A1

6 marks total

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